In a children's park, there is a slide which has a total length of 10 m and a height of 8.0 m (Figure ). A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of the weight. Find (a) the work done by the ladder on the boy as he goes up, (b) the work done by the slide on the boy as he comes down. (c) Find the work done by forces inside the body of the boy.
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"
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Here Given in the question :-
Length l = 10 m
Height h = 8 m
mg = 200 N
(a) When the boy moves up then W.D by the boy against the gravity.
At the time ladder doesn't do any work, so that's zero.
(b) Through friction slides work on the boy
Here Frictional force = (3/10)th is of the weight.
= (3/10) × 200 N
= 60 N
Length of the slide = 10 m
W = f × l
Then work done = 60 N × (-10) m
= -600 J
(Here - sign done due to the movement against the force)
(c) Work done by the forces inside the body of the boy
Here, work is done against the gravity which is
W= m ×g× h
= 200 N × 8 m
= 1600 J .
Hence the W.D by the forces inside the Body of the boy is 1600 J.
Hope it Helps.
Length l = 10 m
Height h = 8 m
mg = 200 N
(a) When the boy moves up then W.D by the boy against the gravity.
At the time ladder doesn't do any work, so that's zero.
(b) Through friction slides work on the boy
Here Frictional force = (3/10)th is of the weight.
= (3/10) × 200 N
= 60 N
Length of the slide = 10 m
W = f × l
Then work done = 60 N × (-10) m
= -600 J
(Here - sign done due to the movement against the force)
(c) Work done by the forces inside the body of the boy
Here, work is done against the gravity which is
W= m ×g× h
= 200 N × 8 m
= 1600 J .
Hence the W.D by the forces inside the Body of the boy is 1600 J.
Hope it Helps.
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8
Answer:
the answer to this a is
1.zero
2. -600J
3.1600J
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